If we consider 2 linearly independent basis as follow: $$\{ |\psi_1 \rangle , |\psi_2 \rangle ... |\psi_n \rangle\}$$ $$\{ |\phi_1 \rangle , |\phi_2 \rangle ... |\phi_n \rangle\}$$

And they are related by a unitary tranform such that: $$U|\psi_i\rangle = |\phi_n\rangle$$

If O is an oprator in basis $\{ |\psi_1 \rangle , |\psi_2 \rangle ... |\psi_n \rangle\}$ and O' is operator in basis $\{ |\phi_1 \rangle , |\phi_2 \rangle ... |\phi_n \rangle\}$, and we could write the spectral decomposition as follows:

$$O = \sum_n \lambda_n |\psi_n \rangle \langle\psi_n|$$ $$O' = \sum_n \mu_n |\phi_n \rangle \langle\phi_n|$$

Would it be true to say that the eigenvalues $\mu_n$ and $\lambda_n$ are equal to one another?

  • $\begingroup$ Hi! What do you think about this? Please share the concepts you are thinking about. Also, are $O$ and $O'$ the same operator in two different bases? Or are they two different operators? $\endgroup$ Commented Nov 6, 2020 at 22:06
  • $\begingroup$ Well the problem I am trying to solve doesn't explicitly mention whether it is the same operator in a different basis however I suspect the relationship that $U|\psi_i\rangle = |\phi_n\rangle$ would suggest that is probably the case. In which case, if I had to guess I would say that the eigenvalues probably are the same since obviously a measured observable would be the same independent of the basis that the operator is represented in. $\endgroup$
    – DJA
    Commented Nov 6, 2020 at 22:15
  • $\begingroup$ Your guess is obviously right: take $O''$ diagonal in the same basis as $O'$, but with different eigenvalues $\mu_n''$ than $O'$. So, now $\lambda_n$ cannot be both equal to $\mu_n$ and $\mu_n''$. It is equal to the respective eigenvalue of the unitarily equivalent operator $O'$. $\endgroup$ Commented Nov 7, 2020 at 17:20

2 Answers 2


You probably mean that, since $\langle \psi_n|U^\dagger =\langle \phi_n|$, $$ O'=UOU^\dagger = \sum_n \lambda_n U|\psi_n \rangle \langle\psi_n| U^\dagger= \sum_n \lambda_n |\phi_n \rangle \langle\phi_n|= O' ,$$ unitarily equivalent.

So, if $O$ and $O'$ are unitarily equivalent, the corresponding transformation of their eigenstate decompositions forces $\lambda_n=\mu_n$.

If they are not unitarily equivalent, the eigenvalues won't coincide: any operator can be written in any unitarily equivalent basis, and the basis by itself cannot tell it what it is or isn't!

Try a 2x2 matrix example. Note the trace and determinant of two unitarily equivalent matrices coincide.

  • $\begingroup$ Could you elaborate on this please? $\endgroup$
    – DJA
    Commented Nov 7, 2020 at 0:05
  • 1
    $\begingroup$ I don't see how your last equality follows. $\endgroup$ Commented Nov 7, 2020 at 4:25
  • $\begingroup$ The question title. $\endgroup$ Commented Nov 7, 2020 at 10:55
  • $\begingroup$ @DJA perhaps you'll find my proof a bit more clear, but also more rote. :) $\endgroup$ Commented Nov 7, 2020 at 17:07
  • $\begingroup$ @DJA sorry there was bad typos in the first version of my answer. Fixed now though. $\endgroup$ Commented Nov 7, 2020 at 21:36

It is given that $$U|\psi_n\rangle = |\phi_n\rangle$$ where $U$ is a unitary matrix and $|\psi_n\rangle$ are eigenkets of $\hat{O}$ and $|\phi_n\rangle$ are eigenkets of $\hat{O'}$ with eigenvalues $\lambda_n$ and $\mu_n$, respectively.

Equivalently, $$|\psi_n\rangle = U^\dagger|\phi_n\rangle$$

Since $\hat{O'}|\phi_n\rangle = \mu_n|\phi_n\rangle$ and $\hat{O}|\psi_n\rangle = \lambda_n |\psi_n\rangle$, and since $\hat{O'} = (U\hat{O}U^\dagger)$, then

$$ \mu_n|\phi_n\rangle = \hat{O'}|\phi_n\rangle = \hat{O'} U|\psi_n\rangle = (U\hat{O}U^\dagger)U|\psi_n\rangle = U\hat{O}|\psi_n\rangle = U\lambda_n|\psi_n\rangle = \lambda_n U|\psi_n\rangle = \lambda_n |\phi_n\rangle$$

Thus, $\lambda_n = \mu_n$ and they share the same eigenvalues.


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